Question: Solve for $x$, ignoring any extraneous solutions: $\dfrac{x^2 + 2}{x + 3} = \dfrac{-x + 8}{x + 3}$
Multiply both sides by $x + 3$ $ \dfrac{x^2 + 2}{x + 3} (x + 3) = \dfrac{-x + 8}{x + 3} (x + 3)$ $ x^2 + 2 = -x + 8$ Subtract $-x + 8$ from both sides: $ x^2 + 2 - (-x + 8) = -x + 8 - (-x + 8)$ $ x^2 + 2 + x - 8 = 0$ $ x^2 - 6 + x = 0$ Factor the expression: $ (x + 3)(x - 2) = 0$ Therefore $x = -3$ or $x = 2$ However, the original expression is undefined when $x = -3$. Therefore, the only solution is $x = 2$.